Question 1
Question 1
∑
(a) State precisely what it means for a to diverge to +∞. [3 marks]
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(b) A sequence has infinitely many positive and negative terms. The series of all positive
terms diverges to +∞, while the series of all negative terms converges. Prove that
every rearrangement of the original series diverges to +∞. [7 marks]
∑ ∑ ∑ √
(c) Let a > 0, a < ∞, and r = ∞ a . Prove that a / r converges. [4 marks]
n n n j=n j n n
∑
(d) With the same notation, prove that a /r diverges. [6 marks]
n n
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Worked solution and marking guidance
Question 1 - Complete course-method solution
Historical official basis: 2023 Q2(a); 2022 Q2(b).
Course-method evidence: official solutions and marker comments in the uploaded 2020–
2026 papers; method families: rearrangement; tail_series; series_partial_sums.
Recognition signal
For rearrangements return to partial sums; for a tail r first write a = r − r ; a square-root denominator suggests rationalisation and telescoping; for a /r
n n n n+1 n n
use blocks in the Cauchy criterion.
First key step
Begin the definition with “all sufficiently late partial sums”; for a tail, first write the difference identity.
(a) For every M > 0 there exists N such that every partial sum with index K ≥ N is greater
than M .
(b) Let the negative subseries have sum A < 0. Given M , take finitely many positive terms
whose sum exceeds M − A. Any rearrangement eventually contains all of these terms. The
negative terms included up to any point have sum at least A. Therefore every sufficiently
late partial sum exceeds M .
(c) Since a = r − r ,
n n n+1
√ √
a √ √ r + r √ √
√n = ( r − r ) n √ n+1 ≤ 2( r − r ).
r n n+1 r n n+1
n n
Compare with the telescoping series.
(d) Fix M . Choose N > M so that r < r /2. Then
N+1 M
∑N
a 1
∑N
r 1
n ≥ a = 1 − N+1 > .
r r n r 2
n M M
n=M n=M
Thus the series fails the Cauchy criterion.
Marking points
Definition: 3 marks; rearrangement: 7; square-root tail: 4; Cauchy divergence: 6.
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MATH40002 Analysis 1 REVISED HISTORICAL COVERAGE
Common errors
Treating an infinite sum as a number whose terms may be rearranged freely; reversing the inequality between negative partial sums and their limit; trying termwise comparison with the harmonic series in (d).
Final self-check
r decreases; any finite sum of negative terms is at least the sum of all negative terms.
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Tony study template card
Recognise the signal → write the first key step → check the theorem hypotheses → complete the course-method sequence → verify using the self-check.
Even if you cannot finish, write the definition, key theorem and first step to earn method marks.
Question 2
Question 2
(a) Classify each set as open, closed, both, or neither: (0, 1), Z, Q, and {0} ∪ [2, ∞). Justify
briefly. [5 marks]
(b) Let f : R → R be continuous and strictly increasing. Prove that f (U ) is open whenever
U ⊂ R is open. [5 marks]
(c) Let f : [1, ∞) → R be uniformly continuous and f (1) = 0. Prove that f (x)/x is bounded.
[5 marks]
(d) Prove that g(x) = sin(x2) is not uniformly continuous on R. [5 marks]
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Worked solution and marking guidance
Question 2 - Complete course-method solution
Historical official basis: 2021 Q4(c); 2022 Q4(c); 2024 Q5(a).
Course-method evidence: official solutions and marker comments in the uploaded 2020–
2026 papers; method families: topology_open_closed; uniform_continuity; IVT_fixed_point.
Recognition signal
Use neighbourhoods/sequences for open and closed sets; use local endpoints + IVT for openness of the image under a continuous strictly monotone map; subdivide and add estimates for uniform continuity on long intervals; disprove
uniform continuity using two sequences.
First key step
For openness of the image, fix y = f (x); for a growth bound, first take ε = 1; for oscillation, choose a phase difference of π.
(a) (0, 1) is open; Z and {0}∪[2, ∞) are closed; Q is neither. Use neighbourhoods and sequential
closedness/density.
(b) If y = f (x) ∈ f (U ), choose δ > 0 with (x − δ, x + δ) ⊂ U . Strict increase gives
f (x − δ/2) < y < f (x + δ/2).
By IVT the entire interval between these values lies in f (U ), so y is interior.
(c) For ε = 1, choose δ ∈ (0, 1) such that |u − v| < δ implies |f (u) − f (v)| < 1. Divide [1, x] into at
most 1 + (x − 1)/δ subintervals of length below δ. Telescoping gives |f (x)| ≤ 1 + (x − 1)/δ, hence
|f (x)|/x ≤ 1 + 1/δ.
√ √
(d) Let x = 2πn + π/2 and y = 2πn + 3π/2. Then |x − y | = π/(x + y ) → 0, while g(x ) = 1
n n n n n n n
and g(y ) = −1.
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Marking points
Set classification: 5 marks; open mapping: 5; linear growth bound: 5; two-sequence method: 5.
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MATH40002 Analysis 1 REVISED HISTORICAL COVERAGE
Common errors
Assuming every continuous function is an open map; incorrectly making the number of subintervals independent of x; choosing two sequences with fixed phase difference but whose distance does not tend to zero.
Final self-check
Use both strict monotonicity and IVT; simplify x − y using a difference of squares.
n n
Tony study template card
Recognise the signal → write the first key step → check the theorem hypotheses → complete the course-method sequence → verify using the self-check.
Even if you cannot finish, write the definition, key theorem and first step to earn method marks.
Question 3
Question 3
′
(a) Give examples of differentiable functions with each property: (i) f (0) = 0 but no
local extremum at 0; (ii) f ′ (x) → 0 but f (x) → ∞; (iii) f bounded but f ′ unbounded. [6
marks]
(b) If f has n + 1 distinct points with the same function value, prove that f (n) vanishes
somewhere, assuming all required derivatives exist. [4 marks]
(c) Use the Mean Value Theorem to prove: (i) f ′ ≡ c ⇒ f (x) = cx + f (0); (ii) if f (x) = 3x
has two distinct solutions, then f ′ (x) = 3 has a solution; (iii) x/(1 + x2) < arctan x < x
for x > 0. [6 marks]
(d) For F (x) = 1/(1 − x3) on (−1, 1), compute F (12)(0)/12! and prove [F (0) + F (x)]/2 ≥ F (x/2)
for 0 < x < 1. [4 marks]
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Worked solution and marking guidance
Question 3 - Complete course-method solution
Historical official basis: 2021 Q5; 2022 Q5(a); 2022 Q5(c).
Course-method evidence: official solutions and marker comments in the uploaded 2020–
2026 papers; method families: MVT_Rolle; convexity; Taylor.
Recognition signal
Use a standard collection for examples; apply Rolle repeatedly to multiple equal-value points; first construct a difference function for MVT; read Taylor coefficients from the power series, and use the second
derivative sign for convexity.
First key step
Start with the simplest example; track the decreasing number of zeros under Rolle; rewrite two solutions of an equation as two zeros of a difference function.
(a) Examples: x3; log x on (0, ∞); sin(x2).
(b) Apply Rolle between consecutive equal-value points to obtain n zeros of f ′, then repeat.
After n applications, f (n) has a zero.
(c)(i) Apply MVT to [0, x]. (ii) Apply MVT to g(x) = f (x) − 3x between two zeros. (iii) MVT
for arctan on [0, x] gives arctan x = x/(1 + t2) for some 0 < t < x.
∑
(d) Since F (x) = ∞ x3m, the coefficient of x12 is 1, so the requested value is 1. Also
m=0
6x(1 + 2x3)
F ′′(x) = > 0 (0 < x < 1),
(1 − x3)3
so convexity gives F (x/2) ≤ [F (0) + F (x)]/2.
Marking points
Examples: 6 marks; repeated Rolle: 4; three MVT applications: 6; power series + convexity: 4.
Common errors
Using a constant function as an example with “no local extrema”; applying Rolle once too few; reversing the arctan inequality; writing the convexity midpoint inequality in the wrong
direction.
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MATH40002 Analysis 1 REVISED HISTORICAL COVERAGE
Final self-check
F should be convex on (0, 1), so its midpoint value does not exceed the average of the endpoint values.
Tony study template card
Recognise the signal → write the first key step → check the theorem hypotheses → complete the course-method sequence → verify using the self-check.
Even if you cannot finish, write the definition, key theorem and first step to earn method marks.
Question 4
Question 4
(a) Let f be twice differentiable and satisfy f
′′
+gf
′−f
= 0 on [a, b], where g is any function.
If f (a) = f (b) = 0, prove f ≡ 0. [5 marks]
∫
b
(b) Let f be continuous on [a, b] and suppose f (x)h(x) dx = 0 for every continuous h
a
with h(a) = h(b) = 0. Prove f ≡ 0. [6 marks]
(c) For continuous f , prove
∫ ∫ (∫ )
x x u
(x − t)f (t) dt = f (t) dt du.
0 0 0
[5 marks]
(d) Let p(0) = 1 and p(x) = 0 for x ̸= 0 on [−1, 1]. Prove p is Darboux integrable and
compute its integral. [4 marks]
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Worked solution and marking guidance
Question 4 - Complete course-method solution
Historical official basis: 2023 Q6.
Course-method evidence: official solutions and marker comments in the uploaded 2020–
2026 papers; method families: ODE_max_principle; functional_test_integral; FTC_integral_identity;
Darboux_sums.
Recognition signal
For an ODE with zero endpoints, use a positive maximum/negative minimum; if the integral vanishes for all test functions, use a local tent function; for nested integral identities,
compare derivatives and initial values; isolate a single-point spike with a narrow partition interval.
First key step
Assume a positive value exists and take a global maximum; explicitly choose a nonnegative test function that is locally positive; name both sides of the identity.
(a) If f has a positive interior maximum x , then f ′(x ) = 0, f ′′(x ) ≤ 0, but the equation gives
0 0 0
f ′′(x ) = f (x ) > 0, contradiction. Thus f ≤ 0. Apply the same argument to −f to get f ≥ 0.
0 0
(b) If f (y) > 0 at an interior point, continuity makes f positive on a small interval. Choose a
continuous non-negative tent function h, zero at the endpoints and positive on that interval;
∫
then f h > 0, contradiction. Treat f (y) < 0 similarly. Endpoint values follow by continuity.
∫ ∫
(c) Let F (x) = x f (t)dt. The right side is G(x) = x F (u)du. The left side is H(x) = xF (x) −
∫ 0 0
x tf (t)dt. By FTC and the product rule, G′ = F = H′, and both vanish at 0.
0
(d) For P = {−1, −δ, δ, 1}, L(p, P ) = 0 and U (p, P ) = 2δ. Hence integrable with integral 0.
δ δ δ
Marking points
Maximum principle: 5 marks; test functions: 6; integral identity: 5; integrability of a spike: 4.
Common errors
Forgetting the zero endpoints when considering where the maximum occurs; choosing a test function that fails the endpoint conditions; exchanging integration order without stating the conditions;
incorrectly calculating the spike's upper sum as 1.
Final self-check
Repeat the argument for −f to obtain equality to 0; both sides of the identity vanish at x = 0.
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MATH40002 Analysis 1 REVISED HISTORICAL COVERAGE
Tony study template card
Recognise the signal → write the first key step → check the theorem hypotheses → complete the course-method sequence → verify using the self-check.
Even if you cannot finish, write the definition, key theorem and first step to earn method marks.
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